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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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901 Easy
901 of 975
Application +4 | -1
The solution of the differential equation dydx=x−yx+4y\frac{d y}{d x}=\frac{x-y}{x+4 y} is (where C{C} is the constant integration)
View answer and explanation
Correct option: D
2xy+4y2=x2+C2 x y+4 y^2=x^2+C
Option D (2xy+4y2=x2+C2 x y+4 y^2=x^2+C) is correct because it is the solution obtained after rearranging and integrating the given differential equation.
902 Medium
902 of 975
Application +4 | -1
Integrating factor of (x+2y3)dydx=y2\left( {x + 2{y^3}} \right)\frac{{dy}}{{dx}} = {y^2} is
View answer and explanation
Correct option: A
e(1y){e^{\left( {\frac{1}{y}} \right)}}
Option A (e(1y){e^{\left( {\frac{{1}}{{y}}} \right)}}) is correct because it is the integrating factor found after rewriting the differential equation and applying the integrating factor formula.
903 Medium
903 of 975
Application +4 | -1
The solution of the differential equation (2x−10y3)dydx,y=0\left(2 x-10 y^3\right) \frac{d y}{d x}, y=0 is
View answer and explanation
Correct option: C
xy2−2y5+cx y^2-2 y^5+c
Option C (xy2−2y5+cx y^2-2 y^5+c) is the correct solution obtained by solving the differential equation using an integrating factor.
904 Easy
904 of 975
Application +4 | -1
The integrating factor of the differential equation (1−x2)dydx−xy=1\left(1-x^2\right) \frac{d y}{d x}-x y=1, is
View answer and explanation
Correct option: C
1−x2\sqrt{1-x^2}
Option C (1−x2\sqrt{{1-x^2}}) is correct because it is the integrating factor obtained from e∫Pdxe^{\int P dx} where P is the coefficient of y in the standard form of the linear differential equation.
905 Easy
905 of 975
Application +4 | -1
The solution of the differential equation dydx=yx+ϕ(yx)ϕ′(yx)\frac{{dy}}{{dx}} = \frac{y}{x} + \frac{{\phi \left( {\frac{y}{x}} \right)}}{{\phi '\left( {\frac{y}{x}} \right)}} is
View answer and explanation
Correct option: A
ϕ(yx)=kx\phi \left( {\frac{y}{x}} \right) = kx
Option A, ϕ(yx)=kx\phi \left( {\frac{y}{x}} \right) = kx, is the correct solution because it satisfies the given differential equation after substitution and integration.
906 Easy
906 of 975
Application +4 | -1
The solution of the equation (2+x)dy−y dx=0(2+x) d y-y \, d x=0 represents a curve passing through a fixed point PP, then the area of equilateral triangle with PP as one vertex and x+y=0x+y=0 as its one side, is
View answer and explanation
Correct option: C
23\frac{2}{\sqrt{3}}
Option C (23\frac{2}{\sqrt{3}}) is correct because the area of the equilateral triangle is calculated using the distance from the point P to the line x+y=0 as the altitude.
907 Medium
907 of 975
Application +4 | -1
dydx+y=2e2x\frac{d y}{d x}+y=2 e^{2 x} then yy is
View answer and explanation
Correct option: A
ce⁡−x+23e2x\operatorname{ce}^{-\mathrm{x}}+\frac{2}{3} \mathrm{e}^{2 \mathrm{x}}
Option A (ce−x+23e2xce^{-\mathrm{x}}+\frac{{2}}{{3}} \mathrm{e}^{2 \mathrm{x}}) is correct because it is the general solution to the given first-order linear differential equation.
908 Easy
908 of 975
Application +4 | -1
y=aemx+be−mxy=a e^{m x}+b e^{-m x} satisfies which of the following differential equations?
View answer and explanation
Correct option: D
d2ydx2−m2y=0\frac{d^2 y}{d x^2}-m^2 y=0
Option D (d2ydx2−m2y=0\frac{d^2 y}{d x^2}-m^2 y=0) is correct because it is the differential equation that results from differentiating the given function twice.
909 Medium
909 of 975
Analyzing +4 | -1
If 2(x−ydydx)(x2+y2)=(x2−y2)(y−xdydx),y(1)=02\left(x-\frac{y d y}{d x}\right)\left(x^2+y^2\right)=\left(x^2-y^2\right)\left(y-\frac{x d y}{d x}\right), y(1)=0, then
View answer and explanation
Correct option: C
ln⁡(x2−y2)+tan⁡−1yx=0\ln \left(x^2-y^2\right)+\tan ^{-1} \frac{y}{x}=0
Option C, ln⁡(x2−y2)+tan⁡−1yx=0\ln \left(x^2-y^2\right)+\tan ^{-1} \frac{y}{x}=0, is correct because it is the solution to the given differential equation with the initial condition y(1)=0.
910 Easy
910 of 975
Application +4 | -1
A solution of the differential equation (dydx)2−xdydx+y=0{\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0 is
View answer and explanation
Correct option: C
y=2x−4y = 2x - 4
Option C (y=2x−4y = 2x - 4) is a solution to the given differential equation, which can be verified by substitution and differentiation.
911 Medium
911 of 975
Application +4 | -1
One of the solution of differential equation dydx=yysin⁡y−x\frac{d y}{d x}=\frac{y}{y \sin y-x} is
View answer and explanation
Correct option: A
y(x+cos⁡y)=2022+sin⁡yy(x+\cos y)=2022+\sin y
Option A (y(x+cos⁡y)=2022+sin⁡yy(x+\cos y)=2022+\sin y) is a solution because it satisfies the given differential equation after rearranging and integrating.
912 Medium
912 of 975
Application +4 | -1
The solution of the differential equation xdydx=y(log⁡y−log⁡x+1)x \frac{d y}{d x}=y(\log y-\log x+1) is
View answer and explanation
Correct option: D
log⁡(yx)=cx\log \left(\frac{y}{x}\right)= c x
Option D (log⁡(yx)=cx\log \left(\frac{y}{x}\right)= c x) is the correct solution, found by recognizing the equation as homogeneous and using the appropriate substitution.
913 Easy
913 of 975
Application +4 | -1
If xsin⁡(yx)dy=[ysin⁡(yx)−y]dxx\sin \left( {\frac{y}{x}} \right)dy = \left[ {y\sin \left( {\frac{y}{x}} \right) - y} \right]dx and y(1)=π2y\left( 1 \right) = \frac{\pi }{2}, then the value of cos⁡(yx)\cos \left( {\frac{y}{x}} \right) is equal to
View answer and explanation
Correct option: C
log⁡x\log x
Option C (log⁡x\log x) is correct because it is the value of cos⁡(yx)\cos(\frac{y}{x}) obtained by solving the homogeneous differential equation with the given initial condition.
914 Medium
914 of 975
Application +4 | -1
The solution of the differential equation dydx−ey∣x+ey−x\frac{d y}{d x}-e^{y \mid x}+e^{y-x} is ......
View answer and explanation
Correct option: A
e−y=e−x−ex+ce^{-y}=e^{-x} - e^x+c
Option A (e−y=e−x−ex+ce^{-y} = e^{-x} - e^x + c) is correct because it is the solution obtained by separating variables and integrating the given differential equation.
915 Medium
915 of 975
Application +4 | -1
The solution of y′=(x+y)2,y(0)=0y^{\prime}=(x+y)^2, y(0)=0 is given by
View answer and explanation
Correct option: A
tan⁡x=x+y\tan x=x+y
Option A (tan x = x+y) is correct because it is the solution obtained by substituting u = x+y and integrating the resulting separable equation with the given initial condition.

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916 Easy
916 of 975
Application +4 | -1
Let y=y(x)y = y(x) be the solution of the differential equation, (x2+1)2dydx+2x(x2+1)y=1{({x^2} + 1)^2}\frac{{dy}}{{dx}} + 2x({x^2} + 1)y = 1 such that y(0)=0y(0) = 0. If a y(1)=π32,\sqrt a \,y(1) = \frac{\pi }{{32}}, then the value of ′a′'a' is
View answer and explanation
Correct option: D
116\frac{1}{{16}}
Option D (116\frac{1}{{16}}) is correct because it is the value of 'a' obtained by solving the differential equation and using the given conditions.
917 Easy
917 of 975
Application +4 | -1
The solution of dydx=ex(sin⁡2x+sin⁡2x)y(2log⁡y+1)\frac{d y}{d x}=\frac{e^x\left(\sin ^2 x+\sin 2 x\right)}{y(2 \log y+1)} is
View answer and explanation
Correct option: A
y2(log⁡y)−exsin⁡2x+c=0y^2(\log y)-e^x \sin ^2 x+c=0
Option A (y2(log⁡y)−exsin⁡2x+c=0y^2(\log y)-e^x \sin ^2 x+c=0) is correct because it is the solution obtained by separating variables and integrating by parts.
918 Medium
918 of 975
Application +4 | -1
Solution of the equation xdy=(y+xf(yx)f′(yx))dxx d y=\left(y+x \frac{f\left(\frac{y}{x}\right)}{f^{\prime}\left(\frac{y}{x}\right)}\right) d x
View answer and explanation
Correct option: C
f(yx)=cxf\left(\frac{y}{x}\right)=c x
Option C, f(yx)=cxf\left(\frac{y}{x}\right)=c x, is the correct solution obtained by solving the given homogeneous differential equation using substitution and integration.
919 Medium
919 of 975
Application +4 | -1
The solution of dydx=yx−sin⁡(yx)\frac{d y}{d x}=\frac{y}{x}-\sin \left(\frac{y}{x}\right) is
View answer and explanation
Correct option: C
y=2xtan⁡−1(cx)y=2 x \tan ^{-1}\left(\frac{c}{x}\right)
Option C (y=2xtan⁡−1(cx)y=2 x \tan ^{-1}\left(\frac{c}{x}\right)) is correct because it is the solution obtained by solving the homogeneous differential equation using the substitution y=vxy=vx.
920 Easy
920 of 975
Understanding +4 | -1
The degree of the equation ex+sin⁡(dydx)=3{e^x} + \sin \left( {\frac{{dy}}{{dx}}} \right) = 3 is
View answer and explanation
Correct option: C
Degree is not defined
The correct answer is C (Degree is not defined) because the equation involves a trigonometric function of a derivative, making it non-polynomial in derivatives.
921 Medium
921 of 975
Application +4 | -1
The solution of ye−x/ydx−(xe−x/y+y3)dy=0y e^{-x / y} d x-\left(x e^{-x / y}+y^3\right) d y=0 is
View answer and explanation
Correct option: A
y22+e−x/y=k\frac{y^2}{2}+e^{-x / y}=k
Option A (y22+e−x/y=k\frac{y^2}{2}+e^{-x / y}=k) is the correct solution, found by rearranging the equation and integrating using appropriate substitutions.
922 Medium
922 of 975
Application +4 | -1
A curve passing through origin, all the normals to which pass through (1,2)(1,2) is
View answer and explanation
Correct option: D
a circle with centre (1,2)(1,2)
Option D (a circle with centre (1,2)) is correct because the equation derived from the given conditions represents a circle centered at (1,2).
923 Easy
923 of 975
Understanding +4 | -1
The solution of the differential equation dxx+dyy=0\frac{{dx}}{x} + \frac{{dy}}{y} = 0 is
View answer and explanation
Correct option: A
xy=cxy = c
Option A (xy=cxy = c) is correct because integrating the given differential equation directly leads to this solution.
924 Easy
924 of 975
Understanding +4 | -1
The degree of differential equation d2ydx2+3(dydx)2=xlog⁡d2ydx2\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x \log \frac{d^2 y}{d x^2} is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the degree of the differential equation is not defined since the equation is not a polynomial in all differential coefficients due to the logarithm.
925 Easy
925 of 975
Application +4 | -1
A curve passing through origin, all the normals to which pass through (x0,y0)\left(x_0, y_0\right) is
View answer and explanation
Correct option: B
x2+y2=2(xx0+yy0)x^2+y^2=2\left(x x_0+y y_0\right)
Option B (x2+y2=2(xx0+yy0)x^2+y^2=2\left(x x_0+y y_0\right)) is correct because it represents a curve passing through the origin where all normals pass through (x0, y0).
926 Medium
926 of 975
Application +4 | -1
If the solution curve of the differential equation dydx=x+y−2x−y\frac{d y}{d x}=\frac{x+y-2}{x-y} passes through the points (2,1)(2,1) and (k+1,2),k>0(k+1,2), k>0, then
View answer and explanation
Correct option: A
2tan⁡−1(1k)=log⁡e(k2+1)2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _e\left(k^2+1\right)
Option A (2tan⁡−1(1k)=log⁡e(k2+1)2 \tan^{-1}(\frac{{1}}{{k}}) = \log_e(k^2+1)) is correct because after solving the homogeneous differential equation and applying the given points, this relationship is obtained.
927 Medium
927 of 975
Application +4 | -1
The solution of the differential equation dydx+xy1−x2=xy,(∣x∣<1)\frac{d y}{d x}+\frac{x y}{1-x^2}=x \sqrt{y},(|x|<1) is y=−f(x)3+C(1−x2)14\sqrt{y}=-\frac{f(x)}{3}+C\left(1-x^2\right)^{\frac{1}{4}}, where f(12)=34f\left(\frac{1}{2}\right)=\frac{3}{4} and CC is an arbitrary constant. Then, the value of f(−12)f\left(-\frac{1}{2}\right) is
View answer and explanation
Correct option: B
34\frac{3}{4}
Option B (34\frac{3}{4}) is correct because it is the value of the function f(x)f(x) at x=−12x=-\frac{1}{2}, where f(x)f(x) is derived from solving the given differential equation.
928 Easy
928 of 975
Application +4 | -1
The curve such that the ratio of the subnormal at any point to the sum of its abscissa and ordinate is equal to the ratio of the ordinate of this point to its abscissa is
View answer and explanation
Correct option: D
y=xlog⁡∣Cx∣y=x \log |C x|
Option D (y=xlog⁡∣Cx∣y=x \log |C x|) is correct because it satisfies the given differential equation derived from the ratio of subnormal to the sum of abscissa and ordinate.
929 Easy
929 of 975
Understanding +4 | -1
To reduce the differential equation dydx+P(x).y=Q(x).yn\frac{{dy}}{{dx}} + P\left( x \right).y = Q\left( x \right).{y^n} to the linear form, the substitution is
View answer and explanation
Correct option: B
v=1yn−1v = \frac{1}{{{y^{n - 1}}}}
Option B (v=1yn−1v = \frac{{1}}{{{y^{n - 1}}}}) is correct because this substitution transforms the Bernoulli equation into a linear differential equation.
930 Medium
930 of 975
Application +4 | -1
The solution of y′2xy1−x2=1+xy^{\prime} \frac{2 x y}{1-x^2}=1+x
y∣x=0=1\left.y\right|_{x=0}=1 is given by
View answer and explanation
Correct option: D
y=11−x2(1+x+x22−x33−x44)y=\frac{1}{1-x^2}\left(1+x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}\right)
The correct answer is D (y=11−x2(1+x+x22−x33−x44)y=\frac{{1}}{{1-x^2}}\left(1+x+\frac{{x^2}}{{2}}-\frac{{x^3}}{{3}}-\frac{{x^4}}{{4}}\right)) because it is the solution of the linear differential equation with the given initial condition.
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